HELP with Chemistry Problem: Equilibrium and Decomposition?

Question

At 275 degrees(C), the equilibrium constant for the reaction below is 1.04x10^-2

NH4Cl(s) <--> NH3(g) + HCl(g)

A .980 g sample of the solid NH4Cl is placed in a 1.000 L closed container and is heated to 275 degrees(C).

a. Will the be any decomposition of the NH4Cl when the system comes to equilibrium at 275 degrees(C)? (Compare Q to K)

b. At equilibrium what will be the partial pressures of the products and the total pressure of the gases inside the container?

c. What will be the mass of the solid NH4Cl in the container at equilibrium?

Any direction on how to do this would be simply fantastic!

Answer 1

a. Remember that only gaseous or dissolved species appear in the equilibrium (and Q) expression. You don&#039;t specify if the K is Kc (concentration) or Kp (pressure). I&#039;ll just assume Kp, since it is easier to handle gaseous equilibria that way. The expression for Kp is:

Kp = P(NH3) * P(HCl).

Since initially there there is neither NH3 nor HCl present, then Q = 0 * 0 =  0. Since K &gt; 0 then of course there will be some decomposition of NH4Cl at 275 °C.

b. Again, assuming Kp, and the unit of Kp is atm²:

Kp = P(NH3) * P(HCl) = 1.04x10^-2 atm². Since P(NH3) = P(HCl) then the partial pressures of both are:

P(NH3) = P(HCl) = √(Kp) = 0.102 atm.

The total pressure is then P(NH3) + P(HCl) = 0.102 atm + 0.102 atm = 0.204 atm.

If the given K is Kc, then the concentrations are 0.102 M, and the partial pressures would have to be worked out using the ideal gas law as shown below in c.

c. For either one of the products,

n = PV/RT = (0.102 atm) (1 L) / (0.082 L*atm/K*mol) (548 K)

n = 0.00227 moles gained.

By the stoichiometry of the reaction, NH4Cl lost the same number of moles as the products individually gained. The mass lost was:

(0.00227 moles) (53.5 g/mole) = 0.121 g,

so the amount of NH4Cl left is:

0.980 - 0.121 g = 0.859 g.

I had to make a lot of assumptions to do this, but it is easy to adjust the answers if my assumptions were wrong, just do the same procedures using the correct K and units.